# 8.6: Real Sequences

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

Recall that a sequence is a function with domain $$\mathbb{N}$$ (or $$\mathbb{N}^{+}$$). A real sequence is a real-valued sequence (that is, the range of the sequence is a subset of the real numbers).

DEFINITION. Subsequence Let $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ be a sequence and $$f \in \mathbb{N}^{\mathbb{N}}$$ be a strictly increasing sequence of natural numbers. Then $\left\langle a_{f(n)} \mid n \in \mathbb{N}\right\rangle$ is a subsequence of $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$.

EXAMPLE 8.4. Let $$s$$ be the sequence $\langle 2 n \mid n \in \mathbb{N}\rangle=\langle 0,2,4,6,8, \ldots\rangle .$ Then the sequence $$t$$ given by $\langle 6 n \mid n \in \mathbb{N}\rangle=\langle 0,6,12,18, \ldots\rangle$ is a subsequence of $$s$$. In this example, $$f(n)=3 n$$ is the function that demonstrates that $$t$$ is a subsequence of $$s$$. Another subsequence of $$s$$ is the sequence $\left\langle 2^{5 n+3} \mid n \in \mathbb{N}\right\rangle$ Recall that a sequence $$\left\langle a_{n}\right\rangle$$ is called non-decreasing if $$a_{n+1} \geq a_{n}$$ for all $$n$$. It is called non-increasing if the inequality is reversed. Everything that is true for a non-decreasing sequence is true, with inequalities reversed, for non-increasing sequences (why?), so rather than state everything twice, we can use the word monotonic to mean a sequence that is either non-increasing (everywhere) or non-decreasing.

LEMMA 8.5. Every non-decreasing real sequence $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ that is bounded above converges to its least upper bound. Every non-increasing real sequence that is bounded below converges to its greatest lower bound.

Proof. We shall only prove the first assertion. Let $$M$$ be the least upper bound of $$\left\langle a_{n}\right\rangle$$. Let $$\varepsilon>0$$. Since $$M$$ is the least upper bound, there is $$N \in \mathbb{N}$$ such that, $0<M-a_{N}<\varepsilon .$ Since the sequence is non-decreasing, $(\forall n \geq N) 0<M-a_{n}<\varepsilon .$ Therefore $$M$$ is the limit of the sequence, as desired.

THEOREM 8.6. Bolzano-Weierstrass Theorem Let $$[b, c]$$ be a closed bounded interval of real numbers and $$s=\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ be a sequence of real numbers such that $(\forall n \in \mathbb{N}) a_{n} \in[b, c] .$ Then $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ has a convergent subsequence with limit in $$[b, c]$$. Discussion. We consider a nested sequence of intervals, all of which contain infinitely many elements of the range of the sequence $$s$$, with the radius of the intervals approaching 0 . We construct a subsequence of $$s$$ by sequentially selecting elements in the intersection of the range of $$s$$ and the successive intervals. We then show that the subsequence we construct is convergent.

PROOF. We prove the theorem for the closed unit interval $$[0,1]$$. It is straightforward to generalize this argument to arbitrary closed bounded intervals.

If the range of the sequence is a finite set, then at least one element of the range, $$a_{n}$$, must have an infinite pre-image. The pre-image of $$a_{n}$$ gives a subsequence that converges to $$a_{n}$$. Therefore we assume that the range of the sequence is infinite. Let $$S$$ be the range of the sequence $$\left\langle a_{n}\right\rangle$$.

We define a nested sequence of closed intervals, $$I_{n}=\left\langle\left[b_{n}, c_{n}\right]\right| n \in$$ $$\mathbb{N}\rangle$$ satisfying

(1) $$I_{0}=[0,1]$$

(2) For all $$n \in \mathbb{N}, I_{n+1} \subset I_{n}$$

(3) $$c_{n}-b_{n}=\frac{1}{2^{n}}$$

(4) For all $$n \in \mathbb{N}, I_{n} \cap S$$ is infinite.

Let $$I_{0}=[0,1]$$. Assume that we have $$I_{n}$$ satisfying the conditions above. At least one of the intervals $$\left[b_{n}, b_{n}+\frac{1}{2^{n+1}}\right]$$ and $$\left[b_{n}+\frac{1}{2^{n+1}}, c_{n}\right]$$ must contain infinitely many elements of $$S$$. Let $$I_{n+1}=\left[b_{n}, b_{n}+\frac{1}{2^{n+1}}\right]$$ if the intersection of this set with $$S$$ is infinite; otherwise let $$I_{n+1}=$$ $$\left[b_{n}+\frac{1}{2^{n+1}}, c_{n}\right]$$. Then $$I_{n+1}$$ satisfies the conditions above.

The sequence of left end-points of the intervals $$I_{n},\left\langle b_{n} \mid n \in \mathbb{N}\right\rangle$$ is non-decreasing. The sequence of right endpoints of the intervals $$I_{n}$$, $$\left\langle c_{n} \mid n \in \mathbb{N}\right\rangle$$ is non-increasing. Furthermore, for any $$m, n \in \mathbb{N}$$, $b_{m}<c_{n} .$ The set $$\left\{b_{n} \mid n \in \mathbb{N}\right\}$$ is bounded above, so by the Least Upper Bound Property the set has a least upper bound, $$\beta$$. Similarly the set $$\left\{c_{n} \mid\right.$$ $$n \in \mathbb{N}\}$$ has a greatest lower bound $$\gamma$$. By Lemma $$8.5$$ \begin{aligned} &\lim _{n \rightarrow \infty} b_{n}=\beta \\ &\lim _{n \rightarrow \infty} c_{n}=\gamma . \end{aligned} By the triangle inequality, for any $$n \in \mathbb{N}$$, $|\beta-\gamma| \leq\left|\beta-b_{n}\right|+\left|b_{n}-c_{n}\right|+\left|c_{n}-\gamma\right| .$ All three terms on the right hand side of the inequality tend to 0 as $$n$$ approaches infinity, so for any $$\varepsilon>0$$, $|\beta-\gamma|<\varepsilon .$ Hence $$\beta=\gamma$$.

We now want to define a subsequence that converges to $$\beta$$, by choosing a point in each interval $$I_{n}$$ in turn. Formally we do this by defining $$f \in \mathbb{N}^{\mathbb{N}}$$ recursively by $f(0)=0$ and $$f(n+1)$$ is the least $$k \in \mathbb{N}$$ such that $[k>f(n)] \wedge\left[a_{k} \in I_{n+1}\right] .$ This is well-defined since $$S \cap I_{n+1}$$ is infinite. Then the sequence $$\left\langle a_{f(n)}\right|$$ $$n \in \mathbb{N}\rangle$$ converges to $$\beta$$. To see this, let $$\varepsilon>0$$. For any $$n \in \mathbb{N}$$ such that $$\frac{1}{2^{n}}<\varepsilon$$, $\left|\beta-a_{f(n)}\right|<c_{n}-b_{n}=\frac{1}{2^{n}}<\varepsilon .$ Therefore $$\left\langle a_{f(n)} \mid n \in \mathbb{N}\right\rangle$$ is a convergent subsequence converging to $$\beta$$.

DEFINITION. Cauchy sequence Let $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ be a sequence. The sequence $$\left\langle a_{n}\right\rangle$$ is a Cauchy sequence if $(\forall \varepsilon>0)(\exists N \in \mathbb{N})(\forall m, n \in \mathbb{N})[m, n \geq N] \Rightarrow\left[\left|a_{m}-a_{n}\right|<\varepsilon\right] .$ THEOREM 8.7. A real sequence converges iff it is a Cauchy sequence. PROOF. $$\Rightarrow$$

Let $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ be a sequence of real numbers that converges to $$a \in \mathbb{R}$$.

Let $$\varepsilon>0$$ and $$N \in \mathbb{N}$$ be such that $(\forall n \geq N)\left|a-a_{n}\right|<\frac{\varepsilon}{2} .$ Then for any $$m, n \geq N$$, $\left|a_{n}-a_{m}\right| \leq\left|a_{n}-a\right|+\left|a-a_{m}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .$ Therefore $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ is a Cauchy sequence.

$$\Leftarrow$$

Let $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ be a Cauchy sequence. Then $(\exists N \in \mathbb{N})(\forall m, n>N)\left|a_{n}-a_{m}\right|<1 .$ Every term in the sequence after the $$N^{\text {th }}$$ term is in the $$\varepsilon$$-neighborhood of $$a_{N}$$. So $(\forall n \geq N) a_{n} \in\left[a_{N}-1, a_{N}+1\right] .$ The sequence $$\left\langle a_{n} \mid n \geq N\right\rangle$$ satisfies the hypotheses of the BolzanoWeierstrass Theorem, and thus has a convergent subsequence.

Let $$\left\langle a_{f(n)} \mid n \in \mathbb{N}\right\rangle$$ be a convergent subsequence of $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ converging to $$a \in \mathbb{R}$$. Let $$\varepsilon>0$$. Since $$\left\langle a_{n}\right\rangle$$ is Cauchy, there is $$N_{1}$$ such that $\left(\forall m, n \geq N_{1}\right)\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2} .$ Furthermore, there is $$N_{2} \in \mathbb{N}$$ such that $\left(\forall n \geq N_{2}\right)\left|a_{f(n)}-a\right|<\frac{\varepsilon}{2} .$ Let $$N_{3} \geq N_{1}, f\left(N_{2}\right)$$. Then $$N_{3} \geq N_{2}$$ and $\left(\forall n \geq N_{3}\right)\left|a_{n}-a\right| \leq\left|a_{n}-a_{f(n)}\right|+\left|a_{f(n)}-a\right|<\varepsilon .$ Therefore the sequence $$\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle$$ converges to $$a$$.

Cauchy sequences get at the essence of the order-completeness of the real numbers. A Cauchy sequence of rational numbers need not converge to a rational number. For instance, let $$a$$ be any irrational number, and let $$a_{n}$$ be the decimal approximation of $$a$$ to the $$n^{\text {th }}$$ digit. The sequence $$\left\langle a_{n}\right\rangle$$ is a Cauchy sequence of rational numbers that converges to an irrational number. However if a Cauchy sequence fails to converge in a set of numbers, it is reasonable to say that there is a gap in the set of numbers. The real numbers are defined so that these gaps are filled.

This page titled 8.6: Real Sequences is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Bob Dumas and John E. McCarthy via source content that was edited to the style and standards of the LibreTexts platform.